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Math Help - De.

  1. #1
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    De.

    Hello,

    Given is y^{4}+4y=0 Who can I solove this? Who can do this step by step?

    Greets Thanks .
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  2. #2
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    By the title "DE", is this a differential equation? Does your equation mean y''''+4y=0?
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  3. #3
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    Quote Originally Posted by Bert
    Hello,

    Given is y^{4}+4y=0 Who can I solove this? Who can do this step by step?

    Greets Thanks .
    It should it be
    y^{(4)}+4y=0
    ---

    If,
    y^{(4)}+4y=0
    The charachteristic equation is,
    k^4+4=0
    Thus,
    k^4=-4
    The solutions are,
    k=1+i,1-i,-1+i,-1-i
    Since the solutions are complex conjakates thus the general solution is,
    y=C_1e^x(\cos x+\sin x)+C_2e^x(\cos x-\sin x) +C_3e^{-x}(\cos x+\sin x)+C_4e^{-x}(\cos x-\sin x)
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  4. #4
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    who do you get those complex solotions?
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  5. #5
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    can you put \frac{\sqrt{2}}{2} into the ct?
    Last edited by Bert; June 26th 2006 at 02:32 AM.
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  6. #6
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    Quote Originally Posted by Bert
    who do you get those complex solotions?
    But the rest you understand?
    ----------
    Method 1) (Informal but it works!)
    k^4=-4
    Thus, (square root),
    k^2=\pm 2 i

    If k^2=2i
    Take square roots,
    k=\pm \sqrt{2}\sqrt{i}
    But, \sqrt{i}=\sqrt{2}/2\pm \sqrt{2}/2 iThus,
    k=\pm \sqrt{2} (\sqrt{2}/2\pm\sqrt{2}/2 i)
    There are hence 4 possibilities (based on plus minus signs).
    k=-1-i,1+i,1-i,-1+i

    Finally if,
    k^2=-2i then by the same argument you get the same solutions, as before.
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  7. #7
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    Method 2) (The is the elegant way. [My method if you are curious was to guess and play around with the roots cuz I was too lazy to work it out])

    You have,
    k^4=-4
    Express this in polar form,
    k^4=4(\cos \pi+i\sin \pi)
    Now apply, de Moivre's theorem
    And the roots are,
    4^{1/4} \left( \cos \left( \frac{\pi}{4}+\frac{2\pi k}{4}\right) +i\sin \left( \frac{\pi}{4}+\frac{2\pi k}{4} \right) \right) one for each value of k=0,1,2,3. And 4^{1/4}=(4^{1/2})^{1/2}=2^{1/2}=\sqrt{2}
    You get your solutions.
    \sqrt{2} (\cos \pi/4+i\sin \pi/4)=1+i
    \sqrt{2} (\cos 3\pi/4+i\sin 3\pi/4)=-1+i
    \sqrt{2} (\cos 5\pi/4+i\sin 5\pi/4)=-1-i
    \sqrt{2} (\cos 7\pi/4+i\sin 7\pi/4)=-1+i
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  8. #8
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    O yes thenk you very wel I see Greets.
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